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5.3 COMPLEX IMPEDANCES (複數阻抗 )
Inductance假設通過電感 L的 sinusoidal current 為
)sin()( tIti mL
則電感 L兩端的 voltage 為
dt
tdiLtv L
L
)()( )cos( tLIm ( 亦為 sinusoid)
InductanceVoltage & current 的 phasors 為
90 mL II
mmL VLIV
Current lags the voltage by 90o
)2
( tT
Im
LIm
Ohm’s Law
IRV
What is the relationship between phaor vltage and phasor current? Does it be similar to Ohm’s law?
90)90( mIL
90 mL II
mL LIV
LIL )90( LILj )90sin90(cos90 jLL ( )
90 LLjZ L 稱為 L的阻抗 (impedance)
Phasor voltage = impedance phasor current (similar to Ohm’s law)
LLL IZV
Note the impedance of an inductance is an imaginary number 虛數 (called reactance). Resistance is a real number.
Capacitance假設通過電容 C的 sinusoidal voltage 為
)cos()( tVtv mc
則電容 C兩端的 current 為
dt
tdvCti c
c
)()( )sin( tCVm
1809090 mmc CVCVI
mC VV
9090 mm ICV)180cos(1
Capacitance
CjCj
CCV
V
I
VZ
m
m
c
cc
11
901
90
( )
mC VV90 mc CVI
Current leads the voltage by 90o
Capacitance
Resistance
RR RIV
The current and voltage are in phase. ( 同相位 )
電容阻抗
電感阻抗
電阻
CjZc
1
LjZL
RZR
Current lags voltage by 90o
Current leads voltage by 90o
Current and voltage are in phase
Kirchhoff’s Laws in Phasor FormWe can apply KVL directly to
phasors. The sum of the phasor voltages equals zero for any closed path.KCL-The sum of the phasor currents entering a node must equal the sum of the phasor currents leaving.
5.4 Circuit Analysis with Phasors and Complex
Impedances
Circuit Analysis Using Phasors and Impedances1. Replace the time descriptions of the voltage and current sources with the corresponding phasors. (All of the sources must have the same frequency.)2. Replace inductances by their complex impedances ZL = jωL. Replace capacitances by their complex impedances ZC = 1/(jωC) or –j(1/ωC) . Resistances have impedances equal to their resistances.
3. Analyze the circuit using any of the techniques studied earlier in Chapter 2, performing the calculations with complex arithmetic.
Example 5.4 Steady-state AC Analysis of a Series Circuit
I?VR?VL?VC?
1. Phasors
30100sV
2. Complex impedances
1503.0500 jjLjZ L
501040500
116
jjC
jZC CLeq ZZRZ 10010050150100 jjj
454.141 100
100arctan100100 22 ( )
3. Circuit Analysis
15707.0)4530(4.141
100
454.141
30100VI
eq
S
Z
)15500cos(707.0)( tti
3. Circuit Analysis
157.7015707.0100V IRR
751.106
15707.09015090V
ILILjL
1054.35
15707.090509011
V
IC
IC
jC Note
90,90 jj
Example 5.5 Series and Parallel Combinations of Complex Impedances.
VC?IL?IR?IC?
1. Phasors
9010 sV
2. Complex impedances 1001.01000 jjLjZ L
10010101000
116
jjC
jZC
5050)45sin45(cos71.704571.704501414.0
01
01.001.0
1
)100/(1100/1
1
/1/1
1
jj
jjZRZ
CRC
o
jj
j
jj
4571.7050
50arctan5050
505001.001.0
01.001.0
01.001.0
1
01.001.0
1
22
Or
3. Circuit Analysis
o
j
4571.70
5050
18010
4571.70
4571.709010
5050
4571.709010
5050100
4571.709010
j
jjZZ
ZVV
RCL
RCsC
)1000cos(10)1801000cos(10)( tttvC
o
j
4571.70
5050
1351414.04571.70
9010
5050
9010
5050100
9010
jjjZZ
VI
RCL
s
1801.0100
18010
R
VI CR
901.090100
18010
100
18010
jZ
VI
C
CC
Example 5.6 Steady-State AC Node-Voltage Analysis
v1(t)?
1. Phasors 902)100sin(2 t05.1)100cos(5.1 t
2. Complex impedances 101.0100 jjjwL
5102000100
116
jjwcj
3. Circuit Analysis
KCL 9025
VV
10
V 211
j
Node 1
KCL Node 2 05.1510
122 j
VV
j
V
22.0)2.01.0( 21 jVjVj
5.11.02.0 21 VjVj
2.05
1
5
1j
j
j
jj
∵
23)2.01.0( 1 jVj
7.291.1681405.0
4.07.0
2.01.0
2.01.0
2.01.0
23
2.01.0
23V1
jj
j
j
j
j
j
j
)7.29100cos(1.16)(1 ttv
5.5 Power in AC Circuits
阻抗有實數 ( 電阻 ) 或虛數 ( 電感 , 電容 ),功率有無實虛 ?
ZjXRZ
mm IZ
V
Z
VI Z
VI mm
1. If X=0 實阻抗 ( 純電阻 )
)cos()( tVtv m )cos()( tIti m
)(cos)()()( 2 tIVtitvtp mm
實功率 (real power)
( 參考Ch5.1)
0
,
RZ
0avgp
1. If X=0 實阻抗 ( 純電阻 )
RIR
VP rms
rmsavg
22
222
2222
2
mmmmrmsrmsavg
rmsrmsrmsrms
avg
IVIVIVP
IVRIR
VP
2m
rms
VV
2. If R=0, X>0 虛阻抗 ( 電感性 )
)cos()( tVtv m )sin()90cos()( tItIti mm
)2sin(2
)sin()cos()()()( tIV
ttIVtitvtp mmmm
虛功率 or 無效功率 (Reactive power)
90
,
jXZ
0avgp
3. If R=0, X<0 虛阻抗 ( 電容性 )
)cos()( tVtv m )sin()90cos()( tItIti mm
)2sin(2
)sin()cos()()()( tIV
ttIVtitvtp mmmm
虛功率 or 無效功率 (Reactive power)
90
,
jXZ
0avgp
)cos()( tVtv m
)cos()( tIti m
)2sin()sin(2
)]2cos(1)[cos(2
)sin()cos()sin()(cos)cos(
)cos()cos()(2
tIV
tIV
ttIVtIV
ttIVtp
mmmm
mmmm
mm
Power for a General Load一般狀況 R≠0 且 X≠0, RLC 電路 ( 有實與虛阻抗 ) 9090 , 0 jXRZ
)2sin(2
1)sin()cos( ),2cos(1(
2
1)(cos2 ttttt ( )
))sin()sin()cos()cos()cos(( ttt
)cos(2
)2sin()sin(2
)]2cos(1)[cos(2
10
mm
Tmmmm
IV
dtwtIV
wtIV
T
T
avg dttpT
P0
)(1
積分為 0 積分為 0
2,
2rmsrms
mm II
VV
)cos(rmsrmsavg IVP
)cos(rmsrmsavg IVPP
)cos( 為 power factor, PF ( 功率因子 )
iv 為 power angle ( 功率角 ) 代表電流 lags 電壓的角度
rmsrmsIV 為 apparent power ( 視出功律 )
)sin(rmsrmIVQ 為 reactive power ( 無效功律 or 虛功率 ), 單位為VAR (volt amperes reactive)
為 real power ( 有效功律 or 實功率 ) 單位為 W
單位為 VA(volt-amperes)
AC Power Calculations
cosrmsrmsIVP
cosPF
iv
sinrmsrmsIVQ
(W)
(VAR)
22 QPIV rmsrms (VA)
rmsrms22 IVQP
RIP 2rms
XIQ 2rms
(W)
(VAR)
(VA)
RIRI
IZ
RVIVIP rms
mrms
mrmsrmsrms
2
22cos
Z
Rcos
2m
rms
VV
RIP 2rms
Proof:
Z
VI
Z
VI m
m ,
XIQ rms2
Proof:
XIXI
IZ
XVIVIQ rms
mrms
mrmsrmsrms
2
22sin
0 1
90 0
pf
pf
L
L
Z
Z
若為純電阻性負載,則
若為純電抗性負載,則
RC pf
RL pf
在 負載中,電流領先電壓,負載有一領先的在 負載中,電流落後電壓,負載有一落後的
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